Problem: In the diagram, each of the three identical circles touch the other two.  The circumference of each circle is 36.  What is the perimeter of the shaded region? [asy]

defaultpen(1);

path p = (1, 0){down}..{-dir(30)}dir(-60){dir(30)}..{dir(-30)}((2, 0) + dir(-120)){-dir(-30)}..{up}(1, 0)--cycle;
fill(p, gray(0.75));

draw(unitcircle);
draw(shift(2 * dir(-60)) * unitcircle);
draw(shift(2) * unitcircle);
[/asy]
Explanation: Join the centre of each circle to the centre of the other two. Since each circle touches each of the other two, then these line segments pass through the points where the circles touch, and each is of equal length (that is, is equal to twice the length of the radius of one of the circles). [asy]
import olympiad;
defaultpen(1);

path p = (1, 0){down}..{-dir(30)}dir(-60){dir(30)}..{dir(-30)}((2, 0) + dir(-120)){-dir(-30)}..{up}(1, 0)--cycle;
fill(p, gray(0.75));

draw(unitcircle);
draw(shift(2 * dir(-60)) * unitcircle);
draw(shift(2) * unitcircle);

// Add lines
draw((0, 0)--(2, 0)--(2 * dir(-60))--cycle);

// Add ticks
add(pathticks((0, 0)--(1, 0), s=4)); add(pathticks((1, 0)--(2, 0), s=4));
add(pathticks((0, 0)--dir(-60), s=4)); add(pathticks(dir(-60)--(2 * dir(-60)), s=4));
add(pathticks((2 * dir(-60))--(2 * dir(-60) + dir(60)), s=4)); add(pathticks((2, 0)--(2 * dir(-60) + dir(60)), s=4));

[/asy]

Since each of these line segments have equal length, then the triangle that they form is equilateral, and so each of its angles is equal to $60^\circ$.

Now, the perimeter of the shaded region is equal to the sum of the lengths of the three circular arcs which enclose it.  Each of these arcs is the arc of one of the circles between the points where this circle touches the other two circles.

Thus, each arc is a $60^\circ$ arc of one of the circles (since the radii joining either end of each arc to the centre of its circle form an angle of $60^\circ$), so each arc is $\frac{60^\circ}{360^\circ} = \frac{1}{6}$ of the total circumference of the circle, so each arc has length $\frac{1}{6}(36)=6$.

Therefore, the perimeter of the shaded region is $3(6) = \boxed{18}$.